Difference between revisions of "2008 AMC 10B Problems/Problem 15"

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We know that <math>a,b>0</math> and that <math>b<100</math>.
 
We know that <math>a,b>0</math> and that <math>b<100</math>.
  
We also know that a must be odd, since the right
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We also know that <math>a^2</math> is odd and thus <math>a</math> is odd, since the right side of the equation is odd. <math>2b</math> is even. <math>2b+1</math> is odd.
 
 
side is odd. An odd number (2b) added to a even number is an odd number.
 
  
 
So <math>a=3,5,7,9,11,13</math>, and the answer is <math>\boxed{A}</math>.
 
So <math>a=3,5,7,9,11,13</math>, and the answer is <math>\boxed{A}</math>.
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~qkddud
 
~qkddud
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== Video Solution by OmegaLearn ==
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https://youtu.be/euz1azVKUYs?t=135
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~ pi_is_3.14
  
 
==See also==
 
==See also==
 
{{AMC10 box|year=2008|ab=B|num-b=14|num-a=16}}
 
{{AMC10 box|year=2008|ab=B|num-b=14|num-a=16}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 11:03, 4 November 2022

Problem

How many right triangles have integer leg lengths $a$ and $b$ and a hypotenuse of length $b+1$, where $b<100$?

$\mathrm{(A)}\ 6\qquad\mathrm{(B)}\ 7\qquad\mathrm{(C)}\ 8\qquad\mathrm{(D)}\ 9\qquad\mathrm{(E)}\ 10$

Solution

By the Pythagorean theorem, $a^2+b^2=b^2+2b+1$

This means that $a^2=2b+1$.

We know that $a,b>0$ and that $b<100$.

We also know that $a^2$ is odd and thus $a$ is odd, since the right side of the equation is odd. $2b$ is even. $2b+1$ is odd.

So $a=3,5,7,9,11,13$, and the answer is $\boxed{A}$.


~qkddud

Video Solution by OmegaLearn

https://youtu.be/euz1azVKUYs?t=135

~ pi_is_3.14

See also

2008 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Problem 16
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

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